Is entire function constant when $ |f(z)|\le \log|z|,\ |z|>1$. Let $ f : \mathbb{C} \to \mathbb{C} ,$ entire and $|f(z)|\le \log|z|,\ |z|>1. $
Show that $f$ is constant.
What first comes to mind is Louville's theorem, but log 's problems with analyticity confuse me.
 A: Pick $\epsilon>0$ and let $r=\exp(\epsilon)>1$. Then $|f(z)|\le\epsilon$ for all $z$ with $|z|=r$ by hypothesis. By the maximum modulus principle, $|f(z)|\le \epsilon$ for all $z$ with $|z|<r$, especially for all $z\in\mathbb D$ . Since $\epsilon$ was arbitrary, $f|_{\mathbb D}=0$, hence $f=0$.
A: Let $z\in \mathbb{C}$, with $|z|=1$. Put $a_n=1+1/n$ and $z_n=a_nz$. By your hypothesis, we have $0\leq |f(z_n)|\leq \log a_n$. Hence $|f(z_n|\to 0$ as $n\to \infty$. But $|f(z_n)|\to |f(z)|$. Hence $f(z)=0$ for all $z$ on the unit circle, and $f=0$. 
A: If you use the Cauchy Integral Formula, the proof is very straight forward. Note, for $\forall n\ge 1$,
$$ |f^{(n)}(0)|=\frac{n!}{2\pi}\bigg|\int_{|z|=r}\frac{f(z)}{z^{n+1}}dz\bigg|\le\frac{n!}{2\pi}\frac{\ln r}{r^{n+1}}2\pi r=\frac{n!\ln r}{2\pi r^n}\to0\text{ as }r\to\infty $$
and hene $f\equiv C$ is a constant. Note $|f(z)|\le \ln |z|$ for $|z|>1$, namely, 
$$ |C|\le \ln|z| \text{ for }\forall |z|>1. $$
Letting $|z|\to1$ gives $C=0$, namely, 
$$ f(z)\equiv0.$$
You can use the same way to show a more general result: If there is a constant $\alpha>0$ such that
$$ |f(z)|\le C|z|^\alpha, \forall z\in\mathbb{C}, $$
then $f(z)$ is a polynomial.
